7.7: Kutatua usawa wa busara

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Malengo ya kujifunza

Kutatua usawa wa busara
Tatua usawa na kazi za busara

Kuwa tayari

Kabla ya kuanza, fanya jaribio hili la utayari.

Kupata thamani ya$x-5$ wakati ⓐ$x=6$ ⓑ$x=-3$ ⓒ$x=5$
Kama amekosa tatizo hili, mapitio Mfano 1.2.16.
Tatua:$8-2 x<12$
Ikiwa umekosa tatizo hili, kagua Mfano 2.6.13.
Andika katika nukuu ya muda:$-3 \leq x<5 $
Ikiwa umekosa tatizo hili, kagua Mfano 2.6.4.

Kutatua usawa wa busara

Tulijifunza kutatua kutofautiana kwa mstari baada ya kujifunza kutatua equations linear. Mbinu hizo zilikuwa sawa na ubaguzi mmoja mkubwa. Wakati sisi tulizidisha au kugawanywa na idadi hasi, ishara ya kukosekana kwa usawa ilibadilishwa.

Baada ya kujifunza tu kutatua usawa wa busara, sasa tuko tayari kutatua usawa wa busara. Ukosefu wa usawa wa busara ni usawa ambao una kujieleza kwa busara.

Ukosefu wa usawa

Ukosefu wa usawa wa busara ni usawa ambao una kujieleza kwa busara.

Ukosefu wa usawa kama vile$\quad \dfrac{3}{2 x}>1, \quad \dfrac{2 x}{x-3}<4, \quad \dfrac{2 x-3}{x-6} \geq x,\quad$ na$\quad \dfrac{1}{4}-\dfrac{2}{x^{2}} \leq \dfrac{3}{x}\quad $ ni usawa wa busara kama kila mmoja huwa na kujieleza kwa busara.

Tunapotatua usawa wa busara, tutatumia mbinu nyingi tulizotumia kutatua usawa wa mstari. Sisi hasa lazima tukumbuke kwamba tunapozidisha au kugawanya na idadi hasi, ishara ya usawa inapaswa kurejea.

Tofauti nyingine ni kwamba ni lazima tuangalie kwa makini thamani gani inaweza kufanya usemi wa busara usiojulikana na hivyo lazima iondokewe.

Wakati sisi kutatua equation na matokeo ni$x=3$, tunajua kuna suluhisho moja, ambayo ni 3.

Tunapotatua usawa na matokeo yake ni$x>3$, tunajua kuna ufumbuzi wengi. Sisi grafu matokeo kwa msaada bora kuonyesha ufumbuzi wote, na sisi kuanza na 3. Tatu inakuwa hatua muhimu na kisha tunaamua kama kivuli upande wa kushoto au kulia. Nambari ya haki ya 3 ni kubwa kuliko 3, kwa hiyo tunapiga kivuli kwa haki.

$clipboard_e564fdf3437198939cd8bf44061cfa74e.png$

Ili kutatua usawa wa busara, sisi kwanza lazima tuandike usawa na quotient moja tu upande wa kushoto na 0 upande wa kulia.

Halafu tunaamua pointi muhimu za kutumia kugawanya mstari wa nambari katika vipindi. Hatua muhimu ni namba ambayo hufanya usemi wa busara sifuri au usiojulikana.

Sisi kisha kutathmini mambo ya nambari na denominator, na kupata quotient katika kila muda. Hii itatambua muda, au vipindi, ambavyo vina ufumbuzi wote wa usawa wa busara.

Tunaandika suluhisho katika maelezo ya muda kuwa makini kuamua kama mwisho wa mwisho umejumuishwa.

Mfano$\PageIndex{1}$

Tatua na uandike suluhisho katika maelezo ya muda:$\dfrac{x-1}{x+3} \geq 0$

Suluhisho

Hatua ya 1. Andika usawa kama quotient moja upande wa kushoto na sifuri upande wa kulia.

Ukosefu wetu wa usawa ni katika fomu hii. \[\dfrac{x-1}{x+3} \geq 0 \nonumber \]

Hatua ya 2. Kuamua pointi muhimu-pointi ambapo kujieleza busara itakuwa sifuri au undefined.

Maneno ya busara yatakuwa sifuri wakati nambari ni sifuri. Tangu$x-1=0$ wakati$x=1$, basi 1 ni hatua muhimu.

Maneno ya busara yatakuwa yasiyojulikana wakati denominator ni sifuri. Tangu$x+3=0$ wakati$x=-3$, basi -3 ni hatua muhimu.

Pointi muhimu ni 1 na -3.

Hatua ya 3. Tumia pointi muhimu kugawanya mstari wa nambari katika vipindi.

$clipboard_e431a0043d1c5bb7e4b4d0de15dc59cbc.png$

Mstari wa nambari umegawanywa katika vipindi vitatu:

\[(-\infty,-3) \quad (-3,1) \quad (1,\infty) \nonumber \]

Hatua ya 4. Mtihani thamani katika kila kipindi. Juu ya mstari wa nambari huonyesha ishara ya kila sababu ya kujieleza kwa busara katika kila kipindi. Chini ya mstari wa nambari onyesha ishara ya quotient.

Ili kupata ishara ya kila sababu kwa muda, tunachagua hatua yoyote katika kipindi hicho na kuitumia kama hatua ya mtihani. Hatua yoyote katika kipindi hicho itatoa ishara sawa, hivyo tunaweza kuchagua hatua yoyote kwa muda.

\[\text { Interval }(-\infty,-3) \nonumber \]

Nambari -4 iko katika kipindi$(-\infty,-3)$. Mtihani$x=-4$ katika maneno katika nambari na denominator.

Nambari:

\[\begin{array}{l} {x-1} \\ {-4-1} \\ {-5} \\ {\text {Negative}} \end{array} \nonumber \]

denominator:

\[\begin{array}{l} {x+3} \\ {-4+3} \\ {-1} \\ {\text {Negative}} \end{array} \nonumber \]

Zaidi ya mstari wa nambari, alama jambo$x-1$ hasi na uangalie sababu$x+3$ hasi.

Kwa kuwa hasi iliyogawanywa na hasi ni chanya, alama chanya chanya katika kipindi hicho$(-\infty,-3)$

$clipboard_e4d5ff31fd44e4c2b9e0115f2b58099f8.png$

\[\text {Interval } (-3,1) \nonumber \]

Nambari ya 0 iko katika kipindi$(-3,1)$. Mtihani$x=0$.

Nambari:

\[\begin{array}{l} {x-1} \\ {0-1} \\ {-1} \\ {\text {Negative}} \end{array} \nonumber \]

denominator:

\[\begin{array}{l} {x+3} \\ {0+3} \\ {3} \\ {\text {Positive}} \end{array} \nonumber \]

Zaidi ya mstari wa nambari, alama jambo$x-1$ hasi na alama$x+3$ chanya.

Kwa kuwa hasi imegawanywa na chanya ni hasi, quotient ni alama hasi katika muda$(-3,1)$.

$clipboard_ee9a065f5ae1eb4312d2506f46f91b4f5.png$

\[\text {Interval }(1, \infty) \nonumber \]

Nambari ya 2 iko katika kipindi$(1, \infty)$. Mtihani$x=2$.

Nambari:

\[\begin{array}{l} {x-1} \\ {2-1} \\ {1} \\ {\text {Positive}} \end{array} \nonumber \]

denominator:

\[\begin{array}{l} {x+3} \\ {2+3} \\ {5} \\ {\text {Positive}} \end{array} \nonumber \]

Zaidi ya mstari wa nambari, alama alama$x-1$ nzuri na alama$x+3$ nzuri.

Kwa kuwa chanya kugawanywa na chanya ni chanya, alama chanya chanya katika muda$(1, \infty)$.

$clipboard_e9f05dbda07764ab489a7211f1b930f81.png$

Hatua ya 5. Kuamua vipindi ambapo usawa ni sahihi. Andika suluhisho katika maelezo ya muda.

Tunataka quotient kuwa kubwa kuliko au sawa na sifuri, hivyo idadi katika vipindi$(-\infty,-3)$ na$(1, \infty) $ ni ufumbuzi.

Lakini vipi kuhusu pointi muhimu?

Hatua muhimu$x=-3$ hufanya denominator 0, hivyo ni lazima iondokewe na suluhisho na tunaiweka alama kwa mabano.

Hatua muhimu$x=1$ hufanya maneno yote ya busara 0. Ukosefu wa usawa unahitaji kujieleza kwa busara kuwa kubwa kuliko au sawa na. Kwa hiyo, 1 ni sehemu ya suluhisho na tutaiweka alama kwa bracket.

$clipboard_eff4e23ab54188da0d5229b26f9790bb8.png$

Kumbuka kwamba wakati tuna ufumbuzi linaloundwa na muda zaidi ya moja tunatumia alama muungano$\cup $, kuunganisha vipindi viwili. Suluhisho katika notation ya muda ni$(-\infty,-3) \cup[1, \infty)$.

Zoezi$\PageIndex{1}$

Tatua na uandike suluhisho katika maelezo ya muda:$\dfrac{x-2}{x+4} \geq 0$

Jibu: $(-\infty,-4) \cup[2, \infty)$

Zoezi$\PageIndex{2}$

Tatua na uandike suluhisho katika maelezo ya muda:$\dfrac{x+2}{x-4} \geq 0$

Jibu: $(-\infty,-2] \cup(4, \infty)$

Sisi muhtasari hatua kwa ajili ya kumbukumbu rahisi.

Jinsi ya Kutatua usawa wa busara

Hatua ya 1. Andika usawa kama quotient moja upande wa kushoto na sifuri upande wa kulia.

Hatua ya 2. Kuamua pointi muhimu-pointi ambapo kujieleza busara itakuwa sifuri au undefined.

Hatua ya 3. Tumia pointi muhimu kugawanya mstari wa nambari katika vipindi.

Hatua ya 4. Mtihani thamani katika kila kipindi. Juu ya mstari wa nambari huonyesha ishara ya kila sababu ya nambari na denominator katika kila kipindi. Chini ya mstari wa nambari onyesha ishara ya quotient.

Hatua ya 5. Kuamua vipindi ambapo usawa ni sahihi. Andika suluhisho katika maelezo ya muda.

Mfano unaofuata unahitaji kwamba sisi kwanza kupata usawa wa busara katika fomu sahihi.

Mfano$\PageIndex{2}$

Tatua na uandike suluhisho katika maelezo ya muda:$\dfrac{4 x}{x-6}<1$

Suluhisho

\[\dfrac{4 x}{x-6}<1 \nonumber \]

Ondoa 1 ili kupata sifuri upande wa kulia.

\[\dfrac{4 x}{x-6}-1<0 \nonumber \]

Andika upya 1 kama sehemu kwa kutumia LCD.

\[\dfrac{4 x}{x-6}-\frac{x-6}{x-6}<0 \nonumber \]

Ondoa nambari na uweke tofauti juu ya denominator ya kawaida.

\[\dfrac{4 x-(x-6)}{x-6}<0 \nonumber \]

Kurahisisha.

\[\dfrac{3 x+6}{x-6}<0 \nonumber \]

Fanya namba ili kuonyesha mambo yote.

\[\dfrac{3(x+2)}{x-6}<0 \nonumber \]

Kupata pointi muhimu.

Quotient itakuwa sifuri wakati nambari ni sifuri. Quotient haijulikani wakati denominator ni sifuri.

\[\begin{array}{rlrl} {x+2} & {=0} & {x-6} & {=0} \\ {x} & {=-2} & {x} & {=6} \end{array} \nonumber \]

Tumia pointi muhimu kugawanya mstari wa nambari katika vipindi.

$clipboard_e21abd0b663d106757f6f6e723393df4e.png$

Mtihani thamani katika kila kipindi.

	\((-\infty,-2)\)	\((-2,6)\)	\((6, \infty)\)
\(x+2)\)	\ ((-\ infty, -2)\)” style="wima align:katikati;” class="lt-math-5164"> x+2 -3+2 -1 -	\ (-2,6)\)” style="wima align:katikati;” class="lt-math-5164"> x+2 0+2 2 +	\ (6,\ infty)\)” style="wima align:katikati;” class="lt-math-5164"> x+2 7+2 9 +
\(x-6\)	\ ((-\ infty, -2)\)” style="wima align:katikati;” class="lt-math-5164"> x-6 -3-6 -9 -	\ (-2,6)\)” style="wima align:katikati;” class="lt-math-5164"> x-6 0-6 -6 -	\ (6,\ infty)\)” style="wima align:katikati;” class="lt-math-5164"> x-6 7-6 1 +

$(-\infty,-2)$

$(-2,6)$

$(6, \infty)$

$x+2)$

\ ((-\ infty, -2)\)” style="wima align:katikati;” class="lt-math-5164">

x+2

-3+2

-1

\ (-2,6)\)” style="wima align:katikati;” class="lt-math-5164">

x+2

0+2

\ (6,\ infty)\)” style="wima align:katikati;” class="lt-math-5164">

x+2

7+2

$x-6$

\ ((-\ infty, -2)\)” style="wima align:katikati;” class="lt-math-5164">

x-6

-3-6

-9

\ (-2,6)\)” style="wima align:katikati;” class="lt-math-5164">

x-6

0-6

-6

\ (6,\ infty)\)” style="wima align:katikati;” class="lt-math-5164">

x-6

7-6

Juu ya mstari wa nambari huonyesha ishara ya kila sababu ya kujieleza kwa busara katika kila kipindi. Chini ya mstari wa nambari onyesha ishara ya quotient.

$clipboard_e667e7f41140a947d1fadb213a3b6434f.png$

Kuamua vipindi ambapo usawa ni sahihi. Tunataka quotient kuwa hasi, hivyo suluhisho linajumuisha pointi kati ya -1 na 6. Kwa kuwa usawa ni madhubuti chini ya, mwisho wa mwisho haujumuishwa.

Tunaandika suluhisho katika notation ya muda kama (-1, 6).

Zoezi$\PageIndex{3}$

Tatua na uandike suluhisho katika maelezo ya muda:$\dfrac{3 x}{x-3}<1$.

Jibu: $\left(-\dfrac{3}{2}, 3\right)$

Zoezi$\PageIndex{4}$

Tatua na uandike suluhisho katika maelezo ya muda:$\dfrac{3 x}{x-4}<2$.

Jibu: $(-8,4)$

Katika mfano unaofuata, nambari daima ni chanya, hivyo ishara ya kujieleza kwa busara inategemea ishara ya denominator.

Mfano$\PageIndex{3}$

Tatua na uandike suluhisho katika maelezo ya muda:$\dfrac{5}{x^{2}-2 x-15}>0$.

Suluhisho

Ukosefu wa usawa ni katika fomu sahihi.

\[\dfrac{5}{x^{2}-2 x-15}>0 \nonumber \]

Sababu ya denominator.

\[\dfrac{5}{(x+3)(x-5)}>0 \nonumber \]

Kupata pointi muhimu. Quotient ni 0 wakati nambari ni 0. Kwa kuwa namba daima ni 5, quotient haiwezi kuwa 0.

Quotient itakuwa haijulikani wakati denominator ni sifuri.

\[\begin{aligned} &(x+3)(x-5)=0\\ &x=-3, x=5 \end{aligned} \nonumber \]

Tumia pointi muhimu kugawanya mstari wa nambari katika vipindi.

$clipboard_e021f22ee9a3dcabc95f9915aaa56196b.png$

Maadili ya mtihani katika kila kipindi. Juu ya mstari wa nambari huonyesha ishara ya kila sababu ya denominator katika kila kipindi. Chini ya mstari wa nambari, onyesha ishara ya quotient.

Andika suluhisho katika maelezo ya muda.

\[(-\infty,-3) \cup(5, \infty) \nonumber \]

Exercise $\PageIndex{5}$

Solve and write the solution in interval notation: $\dfrac{1}{x^{2}+2 x-8}>0$.

Answer: $(-\infty,-4) \cup(2, \infty)$

Exercise $\PageIndex{6}$

Solve and write the solution in interval notation: $\dfrac{3}{x^{2}+x-12}>0 $.

Answer: $(-\infty,-4) \cup(3, \infty)$

The next example requires some work to get it into the needed form.

Example $\PageIndex{4}$

Solve and write the solution in interval notation: $\dfrac{1}{3}-\dfrac{2}{x^{2}}<\dfrac{5}{3 x}$.

Solution

\[\dfrac{1}{3}-\dfrac{2}{x^{2}}<\dfrac{5}{3 x} \nonumber \]

Subtract $\dfrac{5}{3 x}$ to get zero on the right.

\[\dfrac{1}{3}-\dfrac{2}{x^{2}}-\dfrac{5}{3 x}<0 \nonumber \]

Rewrite to get each fraction with the LCD

\[\dfrac{1 \cdot x^{2}}{3 \cdot x^{2}}-\dfrac{2 \cdot 3}{x^{2} \cdot 3}-\dfrac{5 \cdot x}{3 x-x}<0 \nonumber \]

Simplify.

\[\dfrac{x^{2}}{3 x^{2}}-\dfrac{6}{3 x^{2}}-\dfrac{5 x}{3 x^{2}}<0 \nonumber \]

Subtract the numerators and place the difference over the common denominator.

\[\dfrac{x^{2}-5 x-6}{3 x^{2}}<0 \nonumber \]

Factor the numerator.

\[\dfrac{(x-6)(x+1)}{3 x^{2}}<0 \nonumber \]

Find the critical points.

\[\begin{array}{rlrl} {3 x^{2}=0} && {x-6=0} && {x+1=0} \\ {x=0} && {x=6} && {x=-1} \end{array} \nonumber \]

Use the critical points to divide the number line into intervals.

$clipboard_e782e665edacc8e08d9dc6bd0db2081e9.png$

Juu ya mstari wa nambari huonyesha ishara ya kila sababu katika kila kipindi. Chini ya mstari wa nambari, onyesha ishara ya quotient.

Tangu, 0 imechukuliwa, suluhisho ni$(-1,0) \cup(0,6)$ vipindi viwili,$(-1,0)$ na$(0,6)$.

Zoezi$\PageIndex{7}$

Tatua na uandike suluhisho katika maelezo ya muda:$\dfrac{1}{2}+\dfrac{4}{x^{2}}<\dfrac{3}{x}$.

Jibu: $(2,4)$

Zoezi$\PageIndex{8}$

Tatua na uandike suluhisho katika maelezo ya muda:$\dfrac{1}{3}+\dfrac{6}{x^{2}}<\dfrac{3}{x}$.

Jibu: $(3,6)$

Tatua Ukosefu wa usawa na Kazi za busara

Wakati wa kufanya kazi na kazi za busara, wakati mwingine ni muhimu kujua wakati kazi ni kubwa kuliko au chini ya thamani fulani. Hii inasababisha usawa wa busara.

Mfano$\PageIndex{5}$

Kutokana na kazi$R(x)=\dfrac{x+3}{x-5}$, kupata maadili ya x kwamba kufanya kazi chini ya au sawa na 0.

Suluhisho

Tunataka kazi kuwa chini ya au sawa na 0.

\[R(x) \leq 0 \nonumber \]

Badilisha kujieleza kwa busara$R(x)$.

\[\dfrac{x+3}{x-5} \leq 0 \quad x \neq 5 \nonumber \]

Kupata pointi muhimu.

\[\begin{array}{rlrl} {x+3=0} && {x-5=0} \\ {x=-3} && {x=5} \end{array} \nonumber \]

Tumia pointi muhimu kugawanya mstari wa nambari katika vipindi.

$clipboard_e76bb40ea8906c703f6c28ecf1b509d58.png$

Maadili ya mtihani katika kila kipindi. Zaidi ya mstari wa nambari, onyesha ishara ya kila sababu katika kila kipindi. Chini ya mstari wa nambari, onyesha ishara ya quotient. Andika suluhisho katika maelezo ya muda. Tangu 5 imetengwa sisi, usiiingize katika kipindi.

\[[-3,5) \nonumber \]

Exercise $\PageIndex{9}$

Given the function $R(x)=\dfrac{x-2}{x+4}$, find the values of $x$ that make the function less than or equal to 0.

Answer: $(-4,2]$

Exercise $\PageIndex{10}$

Given the function $R(x)=\dfrac{x+1}{x-4}$, find the values of $x$ that make the function less than or equal to 0.

Answer: $[-1,4)$

In economics, the function $C(x)$ is used to represent the cost of producing $x$ units of a commodity. The average cost per unit can be found by dividing $C(x)$ by the number of items $x$. Then, the average cost per unit is \(c(x)=\dfrac{C(x)}{x}).

Example $\PageIndex{6}$

The function$C(x)=10 x+3000$ represents the cost to produce $x$, number of items. Find:

The average cost function, $c(x)$
How many items should be produced so that the average cost is less than $40.

Solution

\[C(x)=10 x+3000 \nonumber \]

The average cost function is $c(x)=\dfrac{C(x)}{x})$. To find the average cost function, divide the cost function by $x$.

\[\begin{aligned} &c(x)=\dfrac{C(x)}{x}\\ &c(x)=\dfrac{10 x+3000}{x} \end{aligned} \nonumber \]

The average cost function is $c(x)=\dfrac{10 x+3000}{x} $

We want the function $c(x)$ to be less than 40.

\[c(x)<40 \nonumber \]

Substitute the rational expression forc(x).

\[\dfrac{10 x+3000}{x}<40, \quad x \neq 0 \nonumber \]

Subtract 40 to get 0 on the right.

\[\dfrac{10 x+3000}{x}-40<0 \nonumber \]

Rewrite the left side as one quotient by finding the LCD and performing the subtraction.

\[\begin{aligned} \dfrac{10 x+3000}{x}-40\left(\dfrac{x}{x}\right) &<0\\ \dfrac{10 x+3000}{x}-\dfrac{40 x}{x} &<0\\ \dfrac{10 x+3000-40 x}{x} &<0 \\ \dfrac{-30 x+3000}{x} &<0 \end{aligned} \nonumber \]

Factor the numerator to show all factors.

\[\begin{array}{ll} {\dfrac{-30(x-100)}{x}<0} \\ {-30(x-100)=0} && {x=0} \end{array} \nonumber \]

Find the critical points.

\[\begin{array}{rl} {-30 \neq 0} & {x-100=0} \\ &{x=100} \end{array} \nonumber \]

More than 100 items must be produced to keep the average cost below $40 per item.

Exercise $\PageIndex{11}$

The function$C(x)=20 x+6000$ represents the cost to produce $x$, number of items. Find:

How many items should be produced so that the average cost is less than $60.

Answer

$c(x)=\dfrac{20 x+6000}{x}$
More than 150 items must be produced to keep the average cost below $60 per item.

Exercise $\PageIndex{12}$

The function$C(x)=5 x+900$ represents the cost to produce $x$, number of items. Find:

How many items should be produced so that the average cost is less than $20.

Answer

$c(x)=\dfrac{5 x+900}{x}$
More than 60 items must be produced to keep the average cost below $20 per item.

Ukosefu wa usawa

Mfano\(\PageIndex{1}\)

Zoezi\(\PageIndex{1}\)

Zoezi\(\PageIndex{2}\)

Jinsi ya Kutatua usawa wa busara

Mfano\(\PageIndex{2}\)

Zoezi\(\PageIndex{3}\)

Zoezi\(\PageIndex{4}\)

Mfano\(\PageIndex{3}\)

Exercise \(\PageIndex{5}\)

Exercise \(\PageIndex{6}\)

Example \(\PageIndex{4}\)

Zoezi\(\PageIndex{7}\)

Zoezi\(\PageIndex{8}\)

Mfano\(\PageIndex{5}\)

Exercise \(\PageIndex{9}\)

Exercise \(\PageIndex{10}\)

Example \(\PageIndex{6}\)

Exercise \(\PageIndex{11}\)

Exercise \(\PageIndex{12}\)